Giúp mình gấp vs Cho 1/x+1/y+1/z=1/xyz CMR: A=√(1+x^2)(1+y^2)(1+z^2) thuộc Q 11/09/2021 Bởi Madeline Giúp mình gấp vs Cho 1/x+1/y+1/z=1/xyz CMR: A=√(1+x^2)(1+y^2)(1+z^2) thuộc Q
Ta có: $\quad \dfrac1x +\dfrac1y +\dfrac1z =\dfrac{1}{xyz}$ $\Leftrightarrow \dfrac{xy + yz + zx}{xyz}=\dfrac{1}{xyz}$ $\Leftrightarrow xy + yz + zx = 1$ Khi đó: $\quad \begin{cases}1 + x^2 = xy + yz + zx + x^2\\1 + y^2 = xy + yz + zx + y^2\\1 + z^2 = xy + yz + zx + z^2\end{cases}$ $\Leftrightarrow \begin{cases}1 + x^2 = (x^2 + xy) + (yz + zx)\\1 + y^2= (y^2 + yz) + (zx + xy)\\1 + z^2 = (z^2 + zx) + (xy + yz)\end{cases}$ $\Leftrightarrow \begin{cases}1 + x^2 = x(x+ y) + z(x+y)\\1 + y^2= y(y + z) + x(y+z)\\1 + z^2 = z(z+ x) + y(z+x)\end{cases}$ $\Leftrightarrow \begin{cases}1 +x^2 = (x+y)(z+x)\\1 + y^2= (x+y)(y+z)\\1 + z^2 = (y+z)(z+x)\end{cases}$ Ta được: $\quad A =\sqrt{(1+x^2)(1+y^2)(1+z^2)}$ $\Leftrightarrow A =\sqrt{(x+y)(z+x)(x+y)(y+z)(y+z)(z+x)}$ $\Leftrightarrow A =\sqrt{(x+y)^2(y+z)^2(z+x)^2}$ $\Leftrightarrow A = |(x+y)(y+z)(z+x)|$ $\Rightarrow A\in \Bbb Q$ Bình luận
Ta có:
$\quad \dfrac1x +\dfrac1y +\dfrac1z =\dfrac{1}{xyz}$
$\Leftrightarrow \dfrac{xy + yz + zx}{xyz}=\dfrac{1}{xyz}$
$\Leftrightarrow xy + yz + zx = 1$
Khi đó:
$\quad \begin{cases}1 + x^2 = xy + yz + zx + x^2\\1 + y^2 = xy + yz + zx + y^2\\1 + z^2 = xy + yz + zx + z^2\end{cases}$
$\Leftrightarrow \begin{cases}1 + x^2 = (x^2 + xy) + (yz + zx)\\1 + y^2= (y^2 + yz) + (zx + xy)\\1 + z^2 = (z^2 + zx) + (xy + yz)\end{cases}$
$\Leftrightarrow \begin{cases}1 + x^2 = x(x+ y) + z(x+y)\\1 + y^2= y(y + z) + x(y+z)\\1 + z^2 = z(z+ x) + y(z+x)\end{cases}$
$\Leftrightarrow \begin{cases}1 +x^2 = (x+y)(z+x)\\1 + y^2= (x+y)(y+z)\\1 + z^2 = (y+z)(z+x)\end{cases}$
Ta được:
$\quad A =\sqrt{(1+x^2)(1+y^2)(1+z^2)}$
$\Leftrightarrow A =\sqrt{(x+y)(z+x)(x+y)(y+z)(y+z)(z+x)}$
$\Leftrightarrow A =\sqrt{(x+y)^2(y+z)^2(z+x)^2}$
$\Leftrightarrow A = |(x+y)(y+z)(z+x)|$
$\Rightarrow A\in \Bbb Q$